An equation of the line tangent to the graph of y=(2x+3)/(3x-2) at the point (1,5) is...do I take the derivative of y=(2x+3)/(3x-2) to find the slope and then plug that into y-5=m(x-1)?

I believe the answer is y = -13x + 18. In the question I had, one of the choices was 13x + y = 18, which is what y = -13x + 18 equals. I got -13 for slope, then plugged it in to get y -5 = -13x + 13

check your derivative

I had y' = -13/(3x-2^2
so the slope would be -13

How??????

Y=-13+18

yes you take the derivative which gets you -13/(3x-2)^2

so that means the slope is -13 and you plug the slope and the point (1,5) into the tangent line equation y-y1=m(x-x1) so the final answer will be
13x+y=18

yes

the slope is about 6.28492178, so then plug and chug that into y-y1 = m(x-x1)

i got the same slope of -13. but according to the wks its wrong???

oops i meant *-18

o wait it is -13 hahah im going crazy